MATH 105: Finite Mathematics 9-6: The Normal …math.wallawalla.edu/~duncjo/courses/math105/.../finite_chapter_9-6.pdfBernoulli Probability and the Normal Distribution Properties of

Bernoulli Probability and the Normal Distribution Properties of the Normal Distribution Examples Conclusion

MATH 105: Finite Mathematics9-6: The Normal Distribution

Prof. Jonathan Duncan

Walla Walla College

Winter Quarter, 2006


Outline

1 Bernoulli Probability and the Normal Distribution

2 Properties of the Normal Distribution

3 Examples

4 Conclusion


Outline



3 Examples

4 Conclusion


Graphing Bernoulli Probability Distributions

We start this section by examining the “Probability Distribution”we get from a Bernoulli Process. This is called the BinomialProbability Distribution.

Example

Construct a probability histogram for a Bernoulli process withn = 6 trials where p = 1

2




Example


2r Pr[r ]

0 C(6, 0)“

12

”0 “12

”6≈ 0.01563

1 C(6, 1)“

12

”1 “12

”5≈ 0.09375

2 C(6, 2)“

12

”2 “12

”4≈ 0.23438

3 C(6, 3)“

12

”3 “12

”3≈ 0.31250

4 C(6, 4)“

12

”4 “12

”2≈ 0.23438

5 C(6, 5)“

12

”5 “12

”1≈ 0.09375

6 C(6, 6)“

12

”6 “12

”0≈ 0.01563




Example


2

r Pr[r ]

0 C(6, 0)“

12

”0 “12

”6≈ 0.01563

1 C(6, 1)“

12

”1 “12

”5≈ 0.09375

2 C(6, 2)“

12

”2 “12

”4≈ 0.23438

3 C(6, 3)“

12

”3 “12

”3≈ 0.31250

4 C(6, 4)“

12

”4 “12

”2≈ 0.23438

5 C(6, 5)“

12

”5 “12

”1≈ 0.09375

6 C(6, 6)“

12

”6 “12

”0≈ 0.01563


As n Increases. . .

As n increases, one could almost say the number of successesbecomes more like a continuous variable. The picture for n = 100is shown below.


As n Increases. . .

As n increases, one could almost say the number of successesbecomes more like a continuous variable. The picture for n = 100is shown below.


The Normal Distribution

As n continues to increase, the Binomial Distribution approachesthe Normal Distribution which is shown below.


The Normal Distribution

As n continues to increase, the Binomial Distribution approachesthe Normal Distribution which is shown below.


Outline



3 Examples

4 Conclusion


General Properties

Properties of the Normal Distribution

The following are properties of the Normal Distribution.

Bell shaped

Symmetric about µ

Probability = Area

Area Under Curve = 1

Probability a value lies between a and b is area under curvebetween a and b.

Standard normal distribution has µ = 0 and σ = 1.


General Properties



Bell shaped

Symmetric about µ

Probability = Area





General Properties



Bell shaped

Symmetric about µ

Probability = Area





General Properties



Bell shaped

Symmetric about µ

Probability = Area





General Properties



Bell shaped

Symmetric about µ

Probability = Area





General Properties



Bell shaped

Symmetric about µ

Probability = Area





General Properties



Bell shaped

Symmetric about µ

Probability = Area





Empirical Rule

Empirical Rule

In a normal distribution, approximately

1 68% of outcomes within 1 standard deviation of the mean.

2 95% of outcomes within 2 standard deviations of the mean.

3 99.7% of outcomes within 3 standard deviations of the mean.


Empirical Rule

Empirical Rule

In a normal distribution, approximately

1 68% of outcomes within 1 standard deviation of the mean.

2 95% of outcomes within 2 standard deviations of the mean.

3 99.7% of outcomes within 3 standard deviations of the mean.


Applying the Empirical Rule

Example

A standardized test has a mean score of µ = 125 and a standarddeviation of σ = 12. 1200 students take the test.

1 How many scored between 113 and 137?

2 How many scored above 125?

3 How many students scored less than 89?



Example







Example



0.68× 1200 = 816





Example


1 How many scored between 113 and 137? (816)





Example




0.5× 1200 = 600




Example



2 How many scored above 125? (600)




Example



2 How many scored above 125? (600)


1

2× (1− .997)× 1200 ≈ 2


Other Shapes

As the mean and standard deviation of a normal distributionchanges, so does the shape of the graph.

µ shifts center

σ changes spread


Other Shapes


µ shifts center

σ changes spread


Other Shapes


µ shifts center

σ changes spread


Other Shapes


µ shifts center

σ changes spread


Using the z-score

Since there are infinitely many normal distributions, we will converteach to the standard normal distribution with µ = 0 and σ = 1.

z-scores

If x is a data point in a normal distribution with mean µ andstandard deviation σ, then the z-score for x is:

z =x − µ

σ


Using the z-score


z-scores


z =x − µ

σ


Using the z-score


z-scores


z =x − µ

σ


Using the z-score


z-scores


z =x − µ

σ

Use the table in the back of yourbook to determine the area un-der the curve between the meanand a given z-score.


Outline



3 Examples

4 Conclusion


Height of Fir Trees

Example

The height of fir trees in a certain forest follows a normaldistribution with µ = 240 and σ = 40 inches.


Height of Fir Trees

Example


(a) How many standard deviations away from the mean are treeswhich are 300 and 120 inches tall?


Height of Fir Trees

Example


(a) How many standard deviations away from the mean are treeswhich are 300 and 120 inches tall?

z =300− 240

40= 1.50

z =120− 240

40= −3.00


Height of Fir Trees

Example


(b) What percent of the trees are between 240 and 300 inches tall?


Height of Fir Trees

Example


(b) What percent of the trees are between 240 and 300 inches tall?

µ = 240

z =300− 240

40= 1.50

A = 0.4332⇒ 43.3%


Height of Fir Trees

Example


(c) What percent of the trees are greater than 300 inches tall?


Height of Fir Trees

Example


(c) What percent of the trees are greater than 300 inches tall?

z =300− 240

40= 1.50

A = 1−0.4332 = .0668⇒ 6.7%


Height of Fir Trees

Example


(d) If you choose a tree at random, what is the probability it isbetween 225 and 275 inches tall?


Height of Fir Trees

Example


(d) If you choose a tree at random, what is the probability it isbetween 225 and 275 inches tall?

z =225− 240

40= −0.13

z =275− 240

40= 0.88

A = 0.0517 + 0.3106 = .3623


Approximating the Binomial Distribution

As we saw at the beginning of this section, the normal distributionis shaped much like the binomial distribution. So much so that wecan use the normal distribution to approximate binomialprobabilities.

Normal Approximation to the Binomial Distribution

When n is large and p is close to 12 the normal distribution can be

used to approximate the binomial distribution with

µ = np and σ =√

np(1− p)







µ = np and σ =√

np(1− p)







µ = np and σ =√

np(1− p)

Why would we want to approximate binomial probabilities? Remem-ber the president approval rating poll?


Presidential Survey Question

Example

Suppose that the president has a job approval rating of 55%. If 30people are surveyed, what is the probability that a majorityapprove?


Presidential Survey Question

Example

Suppose that the president has a job approval rating of 55%. If 30people are surveyed, what is the probability that a majorityapprove?

µ = 30(.55) = 16.5

σ =√

30(.55)(.45) = 2.72

z =16− 16.5

2.72= −0.18

A = 0.0714 + 0.5000 = 0.5714


Presidential Survey Question, Part II

Example

In the same survey, find the probability that:

1 Between 18 and 25 people approve.

2 All but 4 people surveyed approve.



Example






Example



z =18− 16.5

2.72= 0.55

z =25− 16.5

2.72= 3.13

A = 0.4990− 0.2088 = .2902




Example






Example




z =4− 16.5

2.72= −4.60

A = 0.4990 + 0.5000 = .0.9990


Outline



3 Examples

4 Conclusion


Important Concepts

Things to Remember from Section 9-6

1 Properties of a normal distribution

2 Finding z-scores with z = z−µσ

3 Finding areas using the standard normal curve table

4 Approximating binomial probabilities using the normaldistribution and in particular, µ = np and σ =

√np(1− p)


Important Concepts






√np(1− p)


Important Concepts






√np(1− p)


Important Concepts






√np(1− p)


Important Concepts






√np(1− p)


Next Time. . .

Next time we will review for Exam II. Exam II will cover sections7-4 through 9-6 in your text.

For next time

Review sections 7-4 through 9-6


Next Time. . .

Next time we will review for Exam II. Exam II will cover sections7-4 through 9-6 in your text.

For next time

Review sections 7-4 through 9-6

MATH 105: Finite Mathematics 9-6: The Normal …math.wallawalla.edu/~duncjo/courses/math105/.../finite_chapter_9-6.pdfBernoulli Probability and the Normal Distribution Properties of

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